1. **Problem statement:** Find the volume generated by rotating the curve $y = -x^2 + 6x - 8$ bounded by $y=0$ about the y-axis. 2. **Rewrite the problem:** We want the volume of the solid formed by revolving the region bounded by $y = -x^2 + 6x - 8$ and $y=0$ around the y-axis. 3. **Find the intersection points:** Solve $-x^2 + 6x - 8 = 0$ to find the limits of integration. $$-x^2 + 6x - 8 = 0 \implies x^2 - 6x + 8 = 0$$ Factor: $$ (x - 2)(x - 4) = 0 \implies x = 2, 4 $$ 4. **Express $x$ as a function of $y$:** $$ y = -x^2 + 6x - 8 $$ Rewrite as: $$ x^2 - 6x + (8 + y) = 0 $$ Solve quadratic in $x$: $$ x = \frac{6 \pm \sqrt{36 - 4(8 + y)}}{2} = 3 \pm \sqrt{4 - y} $$ 5. **Limits for $y$:** Since $y=0$ is boundary, and the parabola opens downward, $y$ ranges from the minimum value to 0. The vertex is at $x=3$, $$ y_{max} = -3^2 + 6 \times 3 - 8 = -9 + 18 - 8 = 1 $$ So $y$ ranges from 0 to 1. 6. **Volume by shell method about y-axis:** Volume $V = 2\pi \int_{x=2}^{4} x \cdot y \, dx = 2\pi \int_{2}^{4} x(-x^2 + 6x - 8) dx$ 7. **Calculate the integral:** $$ V = 2\pi \int_{2}^{4} (-x^3 + 6x^2 - 8x) dx $$ Integrate term by term: $$ \int (-x^3) dx = -\frac{x^4}{4}, \quad \int 6x^2 dx = 2x^3, \quad \int (-8x) dx = -4x^2 $$ So, $$ V = 2\pi \left[-\frac{x^4}{4} + 2x^3 - 4x^2 \right]_{2}^{4} $$ 8. **Evaluate at bounds:** At $x=4$: $$ -\frac{4^4}{4} + 2 \times 4^3 - 4 \times 4^2 = -\frac{256}{4} + 2 \times 64 - 4 \times 16 = -64 + 128 - 64 = 0 $$ At $x=2$: $$ -\frac{2^4}{4} + 2 \times 2^3 - 4 \times 2^2 = -\frac{16}{4} + 2 \times 8 - 4 \times 4 = -4 + 16 - 16 = -4 $$ 9. **Subtract:** $$ 0 - (-4) = 4 $$ 10. **Final volume:** $$ V = 2\pi \times 4 = 8\pi $$ **Answer:** The volume of the solid formed by rotating the region bounded by $y = -x^2 + 6x - 8$ and $y=0$ about the y-axis is $8\pi$.